[proofplan] The antipodal map $a: S^n \to S^n$ negates every coordinate: $a(x) = -x$. We decompose $a$ as the composition of $n + 1$ reflections, where the $j$-th reflection negates the $j$-th coordinate and fixes all others. Each reflection has degree $-1$ by [Reflections Have Degree $-1$](/theorems/2246). The multiplicativity of degree ([Elementary Properties of Degree](/theorems/2245), part 3) then gives $\deg(a) = (-1)^{n+1}$. [/proofplan] [step:Express the antipodal map as a composition of $n + 1$ coordinate reflections] For each $j \in \{1, 2, \ldots, n + 1\}$, define the reflection \begin{align*} r_j: S^n &\to S^n \\ (x_1, \ldots, x_{n+1}) &\mapsto (x_1, \ldots, x_{j-1}, -x_j, x_{j+1}, \ldots, x_{n+1}). \end{align*} Each $r_j$ is the restriction to $S^n$ of the reflection of $\mathbb{R}^{n+1}$ about the coordinate hyperplane $\{x_j = 0\}$. This map is well-defined on $S^n$: if $|x| = 1$, then $|r_j(x)| = 1$. The composition $r_1 \circ r_2 \circ \cdots \circ r_{n+1}$ negates every coordinate: \begin{align*} (r_1 \circ r_2 \circ \cdots \circ r_{n+1})(x_1, \ldots, x_{n+1}) = (-x_1, -x_2, \ldots, -x_{n+1}) = a(x). \end{align*} Therefore $a = r_1 \circ r_2 \circ \cdots \circ r_{n+1}$. [/step] [step:Compute $\deg(a) = (-1)^{n+1}$ using multiplicativity of degree] By [Reflections Have Degree $-1$](/theorems/2246), $\deg(r_j) = -1$ for each $j \in \{1, \ldots, n + 1\}$. By part (3) of the [Elementary Properties of Degree](/theorems/2245), the degree of a composition is the product of the degrees: \begin{align*} \deg(a) = \deg(r_1 \circ r_2 \circ \cdots \circ r_{n+1}) = \deg(r_1) \cdot \deg(r_2) \cdots \deg(r_{n+1}) = (-1)^{n+1}. \end{align*} [guided] The multiplicativity $\deg(f \circ g) = \deg(f) \cdot \deg(g)$ extends by induction to any finite composition: $\deg(f_1 \circ \cdots \circ f_k) = \prod_{i=1}^k \deg(f_i)$. This is because $(f_1 \circ \cdots \circ f_k)_* = (f_1)_* \circ \cdots \circ (f_k)_*$ by functoriality, and each $(f_i)_*$ is multiplication by $\deg(f_i)$ on $H_n(S^n) \cong \mathbb{Z}$. Composing these multiplications yields multiplication by the product $\prod_i \deg(f_i)$. Applied to the $n + 1$ reflections $r_1, \ldots, r_{n+1}$, each of degree $-1$: \begin{align*} \deg(a) = (-1)^{n+1}. \end{align*} Note the parity: on even-dimensional spheres $S^{2k}$, the antipodal map has degree $(-1)^{2k+1} = -1$. On odd-dimensional spheres $S^{2k-1}$, the antipodal map has degree $(-1)^{2k} = +1$. This distinction will be important for the [Hairy Ball Theorem](/theorems/2248). [/guided] [/step]